derivative of rotation matrix I can't find what is wrong with my thinking here.
We have a rotation R with exponential coordinates $v$.
The rotation matrix is given by the exponential map as :
$$R = e^{v_\times} = I + \frac{v_\times}{1!} + \frac{v_\times^2}{2!} + ...$$
This can be written in a closed form using the Rodrigues' formula :
$$R = e^{v_\times} = I + v_\times sin\theta +v_\times^2(1-cos\theta)$$
where $\theta = \vert\vert v \vert\vert$ and $\theta v_\times = $
$$
\begin{matrix}
0 &-v_3 &v_2\\
v_3&0&-v_1\\
-v_2&v_1&0
\end{matrix}
$$
The derivative of this with $v_i$ is given as:
$$ (A)\frac{dr}{dv_i} = cos\theta (v_i/ \theta)v_\times+sin\theta (v_i/\theta) v_\times^2 + (sin\theta/\theta)[e_i-(v_i/\theta)(v/\theta)]_\times + ((1-cos\theta)/\theta)(e_iv^T/\theta + ve_i^T/\theta-2v_ivv^T/(\theta^3))$$
as in https://arxiv.org/pdf/1312.0788.pdf
BUT
If we calculate the derivative as:
$$
(B)\frac{dR}{dv_i} = \frac{de^{v_\times}}{dv_i} = \frac{dv_\times}{dv_i}e^{v_\times} = (e_i)_\times e^{v_\times}
$$
which is
$$
(e_i)_\times e^{v_\times} = (e_i)_\times (I + v_\times sin\theta +v_\times^2(1-cos\theta))
$$
then is it correct ?
If it is correct why is it different from A ? Please, can someone explain this.
 A: The problem in your argument is that in general, for a matrix that depends on the parameter $\lambda$, the derivative is
$$
\frac{{\rm d}}{{\rm d}\lambda}e^{A(\lambda)} = \int_0^1{\rm d}t~e^{(1-t)A}\frac{{\rm d}A(\lambda)}{{\rm d}\lambda}e^{A}
$$
which is a lot more complex than $({\rm d}A/{\rm d}\lambda) e^A$
A: The problem is that in your proposed formula (B) you seem rely on an assumption that 
$$
\frac{d e^{\mathbf{v}_{\times}} }{ d \mathbf{v_{\times}}} =  e^{\mathbf{v}_{\times}} 
$$
This does not really make sense and you use it in your application of the chain rule. A derivative with respect to a matrix is non-trivial and the left hand side of the expression above does not even lead to a $3\times3$ matrix (which is on the right hand side).
This is the reason, I suspect, why the authors of the paper you linked to did not differentiate directly from the exponential and chose instead to work with the Rodrigues formula.
Even if we restrict to differentiating by the scalar $v_{i}$, there are still difficulties with the matrix exponential.
In the case of the exponential of a scalar value $x$ with the expansion $e^{x} = 1 + x + x^{2}/2! + ...$, we obtain nice properties due to the fact that, for a scalar
$$
\frac{d}{d x} \left(
\frac{x^{k}}{k!} 
\right) =
\frac{x^{k-1}}{(k-1)!} 
$$
In the case of matrix powers, this becomes more tricky. Consider the matrix $\mathbf{v}_{\times}$. Writing it out in full, squaring it and differentiating with respect to $v_{i}$ gives
$$
\frac{d}{d \, v_{i}} 
\left(
\mathbf{v_{\times}^{2}}
\right)
=
\mathbf{e_{i}} \mathbf{v}^{T}
+
\mathbf{v} \mathbf{e_{i}}^{T} 
- 2 v_{i} I
$$
which has no simple relationship with $\mathbf{v}_{\times}$. I.e. in general, there is no nice relationship between 
$$
\frac{d}{d x} 
\left(
\frac{\mathbf{v}^{k}}{k!} 
\right) 
\textrm{ and }
\frac{\mathbf{v}^{k-1}}{(k-1)!} 
$$
The expression described in Gallego and Yezzi for the derivative has been widely and successfully used, it might be a little long but its individual steps are all fairly basic so I'd recommend using it instead.
